JEE Main 2023 — Mathematics Algebra

We know the binomial expansion of ${(a+b)}^{n}$(a+b)n is,

$${(a+b)}^{n}=C0n{a}^{n}{b}^{0}+C1n{a}^{n-1}{b}^{1}+C2n{a}^{n-2}{b}^{2}+......+Cnn{a}^{0}{b}^{n}$$(a+b)n=C0nanb0+C1nan−1b1+C2nan−2b2+......+Cnna0bn

Using the above expansion,

$$x={(8\sqrt{3}+13)}^{13}=C013{(8\sqrt{3})}^{13}+C113{(8\sqrt{3})}^{12}{(13)}^{1}+\ldots$$x=(83​+13)13=C013(83​)13+C113(83​)12(13)1+…

Let ${x}^{'}={(8\sqrt{3}-13)}^{13}=C013{(8\sqrt{3})}^{13}-C113{(8\sqrt{3})}^{12}{(13)}^{1}+\ldots$x′=(83​−13)13=C013(83​)13−C113(83​)12(13)1+…

So, $x-{x}^{'}=2[C113\cdot {(8\sqrt{3})}^{12}{(13)}^{1}+C313{(8\sqrt{3})}^{10}\cdot {(13)}^{3}\ldots ]$x−x′=2[C113⋅(83​)12(13)1+C313(83​)10⋅(13)3…]

Therefore, $x-{x}^{'}$x−x′ is an even integer, hence $[x]$[x] is even

Now, $y={(7\sqrt{2}+9)}^{9}=C09{(7\sqrt{2})}^{9}+C19{(7\sqrt{2})}^{8}{(9)}^{1}+C29{(7\sqrt{2})}^{7}{(9)}^{2}+....$y=(72​+9)9=C09(72​)9+C19(72​)8(9)1+C29(72​)7(9)2+....Now let,${y}^{'}=(7\sqrt{2}-9{)}^{9}=C09(7\sqrt{2}{)}^{9}-C19(7\sqrt{2}{)}^{8}(9{)}^{1}+C29{(7\sqrt{2})}^{7}{(9)}^{2}$y′=(72​−9)9=C09(72​)9−C19(72​)8(9)1+C29(72​)7(9)2

So, $y-{y}^{'}=2[C19{(7\sqrt{2})}^{8}{(9)}^{1}+C39{(7\sqrt{2})}^{6}{(9)}^{3}+\ldots ]$y−y′=2[C19(72​)8(9)1+C39(72​)6(9)3+…]

$y-{y}^{'}=$y−y′= Even integer hence $[y]$[y] is even